If a segment has end points, 0, 4 and 3, 4 How long will the segment be after it is rotated 270
clockwise
To determine how long the segment will be after it is rotated 270 degrees clockwise, we need to find the distance between the new endpoints after the rotation.
Before the rotation, the segment has endpoints 0, 4 and 3, 4. Let's label these points as A (0, 4) and B (3, 4).
To rotate a point (x, y) 270 degrees clockwise about the origin, you can use the following rotation matrix:
[x', y'] = [x*cos(theta) - y*sin(theta), x*sin(theta) + y*cos(theta)]
For a rotation of 270 degrees clockwise, the rotation matrix would be:
[x', y'] = [x*cos(270) - y*sin(270), x*sin(270) + y*cos(270)]
Simplifying this expression:
[x', y'] = [-y, x]
Now let's rotate the endpoints of the segment:
- For point A (0, 4), rotating it 270 degrees clockwise gives us (-4, 0).
- For point B (3, 4), rotating it 270 degrees clockwise gives us (-4, 3).
The new endpoints of the segment after the rotation are (-4, 0) and (-4, 3).
To find the length of the segment between these two points, we can use the distance formula:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, x1 = -4, y1 = 0, x2 = -4, and y2 = 3:
distance = sqrt((-4 - (-4))^2 + (3 - 0)^2)
= sqrt(0^2 + 3^2)
= sqrt(0 + 9)
= sqrt(9)
= 3
Therefore, the segment will be 3 units long after it is rotated 270 degrees clockwise.
A square with the side 6 units in length is reflected across the x-axis. How long are the sides in the reflected figure?
When a shape is reflected across the x-axis, the x-coordinates of its vertices are negated while the y-coordinates remain the same.
In this case, the square has side length 6 units. Let's label its vertices as A, B, C, and D, starting from the top-left vertex in clockwise order.
A = (0, 6)
B = (6, 6)
C = (6, 0)
D = (0, 0)
The reflected figure will have new vertices, but their y-coordinates will remain the same and their x-coordinates will be negated.
A' = (0, -6)
B' = (-6, -6)
C' = (-6, 0)
D' = (0, 0)
To find the length of the sides in the reflected figure, we can calculate the distances between the vertices.
The side length of the reflected figure will be equal to the distances between the reflected vertices.
Side A'B': sqrt((0 - 0)^2 + (-6 - (-6))^2) = sqrt(0 + 0) = 0
Side B'C': sqrt((-6 - 6)^2 + (-6 - 0)^2) = sqrt((-12)^2 + (-6)^2) = sqrt(144 + 36) = sqrt(180) ≈ 13.416
Side C'D': sqrt((-6 - 0)^2 + (0 - 0)^2) = sqrt((-6)^2 + 0) = sqrt(36 + 0) = sqrt(36) = 6
Side D'A': sqrt((0 - 0)^2 + (0 - (-6))^2) = sqrt(0 + 36) = sqrt(36) = 6
Therefore, the sides of the reflected figure will have lengths of 0, 13.416, 6, and 6 units.
A triangle with an area of 25 square units is rotated 180 clockwise. What is the area of the rotated figure?
When a shape is rotated 180 degrees clockwise, the area of the rotated figure remains the same as the original figure.
If the original triangle has an area of 25 square units, then the rotated figure will also have an area of 25 square units.